1.00 g of EuCl_(2), is treated with excess of aqueous AgNO_(3) and all the chlorine is recovered as 1.29 g of AgCl. Calculate the atomic weight of Eu. (Cl = 35.5, Ag = 108)

1.00 g of EuCl_(2), is treated with excess of aqueous AgNO_(3) and all

1.	1.00 g of EuCl_(2), is treated with excess of aqueous AgNO_(3) and all the chlorine is recovered as 1.29 g of AgCl. Calculate the atomic weight of Eu. (Cl = 35.5, Ag = 108)
Answer» <html><body><p></p>Solution :`underset(1g)(<a href="https://interviewquestions.tuteehub.com/tag/eu-446864" style="font-weight:bold;" target="_blank" title="Click to know more about EU">EU</a>)CI_(2)+AgNO_(3) rarr underset(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.29 g)(<a href="https://interviewquestions.tuteehub.com/tag/agci-1969988" style="font-weight:bold;" target="_blank" title="Click to know more about AGCI">AGCI</a>)` <br/> Since <a href="https://interviewquestions.tuteehub.com/tag/ci-408488" style="font-weight:bold;" target="_blank" title="Click to know more about CI">CI</a> atoms are conserved applying POAC for CI atoms moles of CI in `EuCI_(2)` = moles of CI in AgCI <br/> `2xx ` moles of `EuCI_(2)=<a href="https://interviewquestions.tuteehub.com/tag/1xx-1804644" style="font-weight:bold;" target="_blank" title="Click to know more about 1XX">1XX</a>` moles of AgCI <br/> `2xx(1)/((x+35.5xx2))=1 xx(1.29)/((108+35.5))`<br/> ( where x = atomic weight of Eu ) <br/> `:. x= 152.48`</body></html>